Find the length of r between 2 points

[sigma]

Let r be the curve r(t) = (t, tsint, tcost) Find the length of r between the points (0,0,0) and (pi, 0, -pi)

Here's what I've done so far
\(\displaystyle \
\L\
\begin{array}{l}
r(t) = (t,t\sin t,t\cos t) \\
r'(t) = (1,\sin t + t\cos t,\cos t - t\sin t) \\
\left\| {r'(t)} \right\| = \sqrt {2 + t^2 } \\
L = \int_?^? {\sqrt {2 + t^2 } } dt \\
2\int_?^? {\sec ^3 } \theta d\theta \\
\sec \theta \tan \theta + \ln \left| {\sec \theta + \tan \theta } \right| + c \\
\frac{1}{{\frac{{\sqrt 2 }}{{\sqrt {2 + t^2 } }}}}\frac{1}{{\sqrt 2 }} + \ln \left| {\frac{1}{{\frac{{\sqrt 2 }}{{\sqrt {2 + t^2 } }}}} + \frac{1}{{\sqrt 2 }}} \right| \\
\end{array}
\\)

And I only have one word at this point. Lost. I put in "?'s" for the intergral because i'm not sure what they are. Are they pi and negative pi? Really lost with this one.[/tex]

 

[galactus]

It can be kinda tricky. Use parts.

\(\displaystyle \L\\\int{sec^{3}x}dx\)

\(\displaystyle \L\\\int{sec^{2}(x)sec(x)}dx\)

\(\displaystyle \L\\\int{(tan^{2}(x)+1)sec(x)}dx\)

\(\displaystyle \L\\\int{tan^{2}(x)sec(x)}dx+\int{sec(x)}dx\)

Let's concentrate on the left integral. sec alone is easy.

Let \(\displaystyle \L\\u=tan(x), \;\ dv=sec(x)tan(x)dx, \;\ du=sec^{2}(x)dx, \;\ v=sec(x)\)

Make the subs:
\(\displaystyle \L\\uv-\int{vdu}\)

\(\displaystyle \L\\\int{sec^{3}(x)}dx=sec(x)tan(x)-\int{sec^{3}(x)}dx+\int{sec(x)}dx\)

Add \(\displaystyle \L\\\int{sec^{3}(x)}dx\) to both sides:

\(\displaystyle \L\\2\int{sec^{3}(x)}dx=sec(x)tan(x)+\int{sec(x)}dx\)

Divide by 2:

\(\displaystyle \L\\\int{sec^{3}(x)}dx=\frac{1}{2}sec(x)tan(x)+\frac{1}{2}ln(sec(x)+tan(x))+C\)

Don't forget about the 2 from the beginning.

\(\displaystyle \L\\sec(x)tan(x)+ln(sec(x)+tan(x))\)

Then make the proper triangle terms to put it in terms of x.

\(\displaystyle \L\\tan(u)=\frac{x}{\sqrt{2}}, \;\ sec(u)=\frac{\sqrt{x^{2}+2}}{\sqrt{2}}\)

\(\displaystyle \L\\\fbox{\frac{x\sqrt{x^{2}+2}}{2}+ln(\sqrt{x^{2}+2}+x)}\)